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Exercise 11.3 · Q3

Q.Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200200 litres and a maximum of 600600 litres, with probability density function f(x)={k200≤x≤6000otherwisef(x)=\begin{cases}k & 200\le x\le600\\ 0 & \text{otherwise}\end{cases}. Find

(i) the value of kk
(ii) the distribution function
(iii) the probability that the daily sales will fall between 300300 litres and 500500 litres.
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ff is a constant kk over an interval of length 400400, so normalisation fixes k=1400k=\tfrac1{400} (the uniform density); the cdf is a straight ramp, and the requested probability is just the sub-interval's length divided by 400400.

Step 1. (i) Solve for kk. ∫200600k dx=k(600−200)=400k=1⇒k=1400\displaystyle\int_{200}^{600}k\,dx=k(600-200)=400k=1\Rightarrow k=\dfrac1{400}.

Step 2. (ii) Build the distribution function. For x<200x<200: F(x)=0F(x)=0. For 200≤x≤600200\le x\le600: F(x)=∫200x1400 du=x−200400F(x)=\displaystyle\int_{200}^{x}\dfrac1{400}\,du=\dfrac{x-200}{400}. For x>600x>600: F(x)=1F(x)=1. …

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